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Old 2nd December 2009, 01:21 PM   #21
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Quote:
Originally Posted by Gordy View Post
1.
Surely cathode 'follows' the grid. Grid rises with AC input signal and so cathode rises (approximately) the same amount also.
After reading the reply of Erik de Best, I gave it some thought, and realised he (and you too) is right.

So i'm gonna try the diode and the resistor both asap.

So far, everybody thanks!

And as anybody finds documentation about the concertina regarding the bias, let me know!
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Old 2nd December 2009, 04:06 PM   #22
Arnulf is offline Arnulf  Europe
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There is no magic and therefore not much to read about, this is just how tubes work !

Previous stage sets grid voltage. B+ and load (combination of both resistors and tube itself) determine current passing through cathodyne.

Let's assume for a moment that cathodyne was just turned on, its B+ at normal operating level. When voltage is applied to grid from the previous stage, tube will begin to conduct. As it begins to conduct, current flows through it and causes voltage drop across load resistors. This means that voltage at the anode and cathode moves away from its starting position of B+ and 0V (towards 1/2 of B+ approximately). The more you open the tube, the more its resistance drops, the more current it conducts and the closer together both anode and cathode voltage move. At some point they just cannot go any closer due to tube's remaining on resistance, let alone pass one another. This is the quiescent point and bias voltage and therefore current is set when difference between grid and cathode voltage becomes large enough to keep current thrugh tube limited. As current is limited, cathode voltage cannot rise any further and because it cannot rise any further with regards to grid voltage (making Vg even more negative), current is limited.

Got it ?
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Old 2nd December 2009, 05:25 PM   #23
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Quote:
Originally Posted by Arnulf View Post
There is no magic and therefore not much to read about, this is just how tubes work !

Previous stage sets grid voltage. B+ and load (combination of both resistors and tube itself) determine current passing through cathodyne.

Let's assume for a moment that cathodyne was just turned on, its B+ at normal operating level. When voltage is applied to grid from the previous stage, tube will begin to conduct. As it begins to conduct, current flows through it and causes voltage drop across load resistors. This means that voltage at the anode and cathode moves away from its starting position of B+ and 0V (towards 1/2 of B+ approximately). The more you open the tube, the more its resistance drops, the more current it conducts and the closer together both anode and cathode voltage move. At some point they just cannot go any closer due to tube's remaining on resistance, let alone pass one another. This is the quiescent point and bias voltage and therefore current is set when difference between grid and cathode voltage becomes large enough to keep current thrugh tube limited. As current is limited, cathode voltage cannot rise any further and because it cannot rise any further with regards to grid voltage (making Vg even more negative), current is limited.

Got it ?
That I get! (and got)

But... assume the concertina with ecc81 I had at first.
On the splitter valve the kathode is 20v higher than the anode.So bias on grid is -20v. voltage between anode and kathode is measured 145v... so according to all the curves the tube should block! Or if you look at the current through the second tube (measured) :7 mA. With 7mA and -20v there should be an anode kathode voltage of over 500v.

On the other hand, the ecc88. Here the difference between the g-k is -3v.
Also about 7mA is passing through the valve. V a-k is 136v. with -3v the V a-k should be according to the curves: 115V. Or: with 136v a-k and a bias Vg of -3v there should run about 15 mA.

Obvious, the curves are worthless to predict Ia or V a-k in case of the splitter stage. Or not? (on the first gain stage they work like a charm)
In both cases, the circuit manages to settle itself on 7 mA, with a 1/3 Vb+ over a-k. Coincidense? Probably not.
How does it do that? No idea...
I would like to understand what happens here...so i'm looking for a read with an explanation, a method to calculate or use curves to accurately design such a stage.

Maybe I'm stupid to not get it right away or find it out on my own.. but I can ask for some help.... can't I?

Last edited by pauldune; 2nd December 2009 at 05:32 PM.
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Old 2nd December 2009, 05:46 PM   #24
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Old 2nd December 2009, 06:08 PM   #25
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Quote:
Originally Posted by aardvarkash10 View Post
Super! This was exactly wat I was looking for!!
It even has the solution for arcing.....

On the downside, It doesnt explain the DC coupled variant so nice as the normal version... But maybe with this explanation, I can find out for myself.

So Thanks!
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Old 2nd December 2009, 06:29 PM   #26
Arnulf is offline Arnulf  Europe
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Quote:
Originally Posted by pauldune View Post
On the splitter valve the kathode is 20v higher than the anode.
This doesn't make any sense. I'm wondering whether you've got your numbers right. If cathodyne tube is in cutoff all the time, your cathodyne isn't working.

Quote:
Obvious, the curves are worthless to predict Ia or V a-k in case of the splitter stage. Or not? (on the first gain stage they work like a charm)
Curves represent an average performance of number of tubes that were produced long time ago. They do not necessarily reflect reality of your particular example but they are usually more reliable than those of semiconductor devices.

Quote:
In both cases, the circuit manages to settle itself on 7 mA, with a 1/3 Vb+ over a-k. Coincidense? Probably not.
No, physics

Quote:
I would like to understand what happens here...so i'm looking for a read with an explanation, a method to calculate or use curves to accurately design such a stage.
This is pretty pointless without schematic and actual measurements but I'll give it another shot regardless.

If tube was a perfect conductor, the current through it would be limited only by B+ / (Ra + Rk). This is the maximum current that can ever pass through the tube. Ergo: draw a load line (with R = Ra + Rk) into tube characteristics graph, with (B+, 0 mA) being one point and (0V, Imax) being the other. Your splitter will operate along this line.

In reality there will be some voltage drop across the tube and this drop is anode to cathode voltage and this voltage is determined by grid to cathode voltage.

So start from cutoff (rightmost point on the loadline): Vk = 0V so Vgk = very large and tube starts conducting as hard as it can. Vk wants to shoot up towards B+/2 but by doing so, it also lowers grid to cathode voltage (moves the point left along the loadline and moves to lower grid curve). Vk cannot go above B+/2 and Vg is held steady as well either by resistors or by the previous stage.

Vg is normally chosen so that it is lower than B+/2 (more like B+/4 but you could put it higher as long as your tube still has enough room to drive Vgk negative enough to prevent grid current from flowing). Eventually Vgk becomes low enough for system to reach equilibrium and it will not move unless Vg changes (externally). If you picked too high a grid voltage so that Vgj never becomes 0V or less, your tube will self-destruct (*).

Since you know both limiting points of Vgk, you can calculate the operating point. Vgkmin will be calculated as above and Vgkmax is usually 0V. Pick a value halfway inbetween these and that's your quiescent grid voltage. Take it to your loadline and determine Va and Ia from that. Then you can determine quiescent voltages at anode and cathode either by:

1: calculating voltage drops from Ia and Ra (Rk) and add/subtract those from supply voltages (0V and B+)

or

2: taking Va reading from the loadline and noting that B+ = 2 * Vk + Va (from loadline), this being true because Ra = Rk, so you can get Vk and Va (actual) from that expression

(*): remark regarding tube destruction assumes that your tube will tolerate ABSOLUTELY NO grid current. This is not true for real life tubes (they will survive some grid current so Vgk can actually be higher than 0V) but you don't want to push them right to the edge anyway. In thery you could have a functional splitter with Vgkmax > 0V and all of the above would still apply.
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Old 2nd December 2009, 06:30 PM   #27
Arnulf is offline Arnulf  Europe
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There is NO DIFFERENCE between DC coupled variant and AC coupled with resistor pair setting Vg as far as your question goes.
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Old 2nd December 2009, 07:00 PM   #28
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Originally Posted by Arnulf View Post
There is NO DIFFERENCE between DC coupled variant and AC coupled with resistor pair setting Vg as far as your question goes.
Thanks Arnulf!
I looked at the valve wizard site, and going to read and try to understand your story...so far its clear to me now.
Only the ECC81 doesnt fit.
Optimizing a 807 PP amplifiers input stage and phase splitter...
Both schematics are in the startpost.
In the first are real measured values. In the ecc88 version are my (wrong) predicted values. Strange thing is, the ECC88 version I build after that is very close to your calculations... considering the CCS.

In real life the ECC88 measures like this:

V1: (gain stage) I a-k=8,2 mA / Vbias =-1,47v / Va= 107V
V2: (splitter) I a-k 7 mA / Vk=110V / V bias= -3v / Va=246v / V a-k= 136v
V++=375V

Total gain is about 30x; Distortion is very low (<0,07%) till 60v pp. Then the 807 starts pulling grid current, and the scope image look exactly like the explanation on the valvewizard site. The cathode fed 807 doesnt distort much, the anode fed does! But i'm not gonna drive it into grid current anyway... The amp outputs more than 10W at this point anyway, and this is more than I will need.
But if I wanted to drive the 807's optimally, I should design a stage wich could deliver 92 V undistorted.Or if I could somehow lower the Bias from -30v to -46v, I think the concertina could drive it undistorted to full output.
I could lower the bias current , but i dont want to leave class A operation so soon....

About the ecc81;you think I mixed some numbers up? Its possible, but fortunately a friend of mine build the same amp 22 years ago, and i'm gonna let him remeasure the voltages.

Last edited by pauldune; 2nd December 2009 at 07:05 PM. Reason: typo
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Old 2nd December 2009, 07:53 PM   #29
Arnulf is offline Arnulf  Europe
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Either that or you've got some really funky tube there. Use another power supply and put -20V on the grid and measure cathode current. If you don't have 20V PSU handy just tie grid to ground and insert 20V zener in series with the ammeter into cathode. I think you're not going to read 7 mA ...

I'm to lazy to read this entire thread again but your ECC88 values look perfectly sane - what's wrong with them in your opinion ? If Vg = 107V, then Vk will be close as well. At Vgk = 0V the loadline says current will be 11 mA and Va-k is , then dropping by ~1 mA for each -1V in Vgk. Since B+/2 is 375/2 (187.5V), Va-k will be somewhere between 35V and 2*(187.5V - 107V) = 161V. Now let's see, Vgk can only vary from -0V to -4.5V to keep Va-k within that range so Vk can only be in range of 107V to 111.5V. Expected current at that anode to cathode voltage can be read from the loadline and it appears that Vgk will be approximately -3.5V, Vk = 110.5V, Va = 264.5V. This is not far from your readings.
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Old 2nd December 2009, 08:03 PM   #30
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Originally Posted by Arnulf View Post
Either that or you've got some really funky tube there. Use another power supply and put -20V on the grid and measure cathode current. If you don't have 20V PSU handy just tie grid to ground and insert 20V zener in series with the ammeter into cathode. I think you're not going to read 7 mA ...

I'm to lazy to read this entire thread again but your ECC88 values look perfectly sane - what's wrong with them in your opinion ?
i tend to type a bit to much... sorry about that. But youre doing a great job!(educating me) I think there's nothing wrong with the ecc88; I put the values there, to compare. Also, because the measurements come close to the calculations, following your methods.

That leaves the ecc81; im going to remeasure before anything else, but it played beautifully with these tube, and I had 4 brands of ecc81's in it. Al more or less the same result.

Anyway, Thanks again.

I just soldered in a 4007, and it seems to fix the arcing problem!

So I'm really happy
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